{"id":40998,"date":"2020-03-05T15:53:11","date_gmt":"2020-03-05T10:23:11","guid":{"rendered":"https:\/\/cracku.in\/blog\/?p=40998"},"modified":"2020-03-05T15:53:11","modified_gmt":"2020-03-05T10:23:11","slug":"advance-maths-questions-for-ssc-cgl-pdf","status":"publish","type":"post","link":"https:\/\/cracku.in\/blog\/advance-maths-questions-for-ssc-cgl-pdf\/","title":{"rendered":"Advance Maths Questions for SSC CGL PDF"},"content":{"rendered":"

Advance Maths Questions for SSC CGL PDF<\/strong><\/span><\/h1>\n

Download SSC CGL Advance Maths Questions PDF based on previous papers very useful for SSC CGL Exams. Top-15 Very Important Maths Questions for SSC CGL Exam.<\/p>\n

Download Advance Maths Questions for SSC CGL PDF <\/a><\/p>\n

Take a free SSC-CGL mock test<\/a><\/p>\n

Get 200 SSC mocks for just Rs. 249. Enroll here<\/a><\/p>\n

Question 1:\u00a0<\/b>A car is travelling on a straight road leading to a tower. From a point at a distance of 500 m from the tower, as seen by the driver, the angle of elevation of the top of the tower is 30\u00b0. After driving towards the tower for 10 seconds, the angle of elevation of the top of the tower as seen by the driver is found to be 60\u00b0. Then the speed of the car is<\/p>\n

a)\u00a0135 km\/hr.<\/p>\n

b)\u00a0110 km\/hr.<\/p>\n

c)\u00a0120 km\/hr.<\/p>\n

d)\u00a090 km\/hr.<\/p>\n

Question 2:\u00a0<\/b>The value of $\\frac{sin39^{\\circ}}{cos51^{\\circ}}$ + $2 tan11^{\\circ} tan31^{\\circ} tan45^{\\circ} tan59^{\\circ} tan79^{\\circ}$ – $3(sin^{2}21^{\\circ} + sin^{2}69^{\\circ})$ is<\/p>\n

a)\u00a01<\/p>\n

b)\u00a02<\/p>\n

c)\u00a0\u00ad-1<\/p>\n

d)\u00a00<\/p>\n

Question 3:\u00a0<\/b>Find equation of the perpendicular to segment joining the points A(0,4) and B(-5,9) and passing through the point P. Point P divides segment AB in the ratio 2:3.<\/p>\n

a)\u00a0x – y = 8<\/p>\n

b)\u00a0x – y = -8<\/p>\n

c)\u00a0x + y = -8<\/p>\n

d)\u00a0x + y = 8<\/p>\n

Question 4:\u00a0<\/b>In a class, there are 40 students. Some of them passed the examination and others failed. Raman\u2019s rank among the student who have passed is 13 th from top and 17 th from bottom. How many students have failed?<\/p>\n

a)\u00a011<\/p>\n

b)\u00a010<\/p>\n

c)\u00a09<\/p>\n

d)\u00a0Cannot be determined<\/p>\n

Question 5:\u00a0<\/b>A number is increased by 30%, then decreased by 30%, then further decreased by 30%. What is the net increase\/decrease percent in the number(correct to the nearest integer)?<\/p>\n

a)\u00a036% decrease<\/p>\n

b)\u00a036% increase<\/p>\n

c)\u00a040% increase<\/p>\n

d)\u00a040% decrease<\/p>\n

Take a free mock test for SSC CGL<\/a><\/p>\n

Download SSC CGL Previous Papers PDF<\/a><\/p>\n

18000+ Questions – Free SSC Study Material<\/a><\/p>\n

Question 6:\u00a0<\/b>A cuboid has dimensions $8 cm \\times 10 cm \\times 12 cm$. It is cut into small cubes of side $2 cm$. What is the percentage increase in the total surface area?<\/p>\n

a)\u00a0286.2<\/p>\n

b)\u00a0314.32<\/p>\n

c)\u00a0250.64<\/p>\n

d)\u00a0386.5<\/p>\n

Question 7:\u00a0<\/b>A scooter is sold at three successive discounts of 10%, 5% and 2%. If the marked price of the scooter is Rs.18,000, find its net selling price.<\/p>\n

a)\u00a0Rs.15028.20<\/p>\n

b)\u00a0Rs.15082.00<\/p>\n

c)\u00a0Rs.15082.20<\/p>\n

d)\u00a0Rs.15080.00<\/p>\n

Question 8:\u00a0<\/b>The average marks obtained by 150 students in an examination is 40. If the average marks of passed students is 60 and that of the failed students is 20, then what is the number of students who passed the examination?<\/p>\n

a)\u00a025<\/p>\n

b)\u00a050<\/p>\n

c)\u00a075<\/p>\n

d)\u00a0100<\/p>\n

Question 9:\u00a0<\/b>A solution of milk and water contains milk and water in the ratio of 3 : 2. Another solution of milk and water contains milk and water in the ratio of 2 : 1. 40 liters of the first solution is mixed with 30 liter of the second solution. The ratio of milk and water in the resultant solution is:<\/p>\n

a)\u00a022:13<\/p>\n

b)\u00a013:22<\/p>\n

c)\u00a06:5<\/p>\n

d)\u00a05:6<\/p>\n

Question 10:\u00a0<\/b>Teena, Reena and Sheena start a business with investment of respectively \u20b9 24000, \u20b9 28000 and \u20b9 20000. Teena invests for 8 months, Reena invest for 10 months and Sheena invests for one year. If the total profit at the end of year is \u20b9 25810, then what is the share of Teena?<\/p>\n

a)\u00a0\u20b96960<\/p>\n

b)\u00a0\u20b910150<\/p>\n

c)\u00a0\u20b97940<\/p>\n

d)\u00a0\u20b98700<\/p>\n

FREE SSC EXAM YOUTUBE VIDEOS<\/a><\/p>\n

More SSC CGL Important Questions and Answers PDF<\/a><\/p>\n

Question 11:\u00a0<\/b>How many job applicants had applied if the ratio of selected to unselected was 19:17. If 1,200 less had applied and 800 less selected, then the ratio of selected to unselected would have been 1:1.<\/p>\n

a)\u00a06000<\/p>\n

b)\u00a07200<\/p>\n

c)\u00a08400<\/p>\n

d)\u00a04800<\/p>\n

Question 12:\u00a0<\/b>A train takes 1 minute to cross a stationary pole. How much time (in seconds) will it take to cross a bridge whose length is twice that ofthe train?<\/p>\n

a)\u00a0120<\/p>\n

b)\u00a090<\/p>\n

c)\u00a0150<\/p>\n

d)\u00a0180<\/p>\n

Question 13:\u00a0<\/b>Pipe P can fill a tank alone in 7 hours. Pipe Q can fill the same tank alone in 13 hours. In how much time can they togetherfill the tank?<\/p>\n

a)\u00a03 hours 18 minutes<\/p>\n

b)\u00a03 hours 45 minutes<\/p>\n

c)\u00a04 hours 12 minutes<\/p>\n

d)\u00a04 hours 33 minutes<\/p>\n

Question 14:\u00a0<\/b>Two places P and Q are 162 km apart. A train leaves P for Q and simultaneously another train leaves Q for P. They meet at the end of 6 hours. If the former train travels 8km\/hour faster then the other, then speed of train from Qis<\/p>\n

a)\u00a09 $\\frac{1}{2}$ km\/hr<\/p>\n

b)\u00a010 $\\frac{5}{6}$ km\/hr<\/p>\n

c)\u00a012\u00a0$\\frac{5}{6}$ km\/hr<\/p>\n

d)\u00a09 $\\frac{1}{2}$ km\/hr<\/p>\n

e)\u00a08\u00a0$\\frac{1}{2}$ km\/hr<\/p>\n

Question 15:\u00a0<\/b>ABCD is a square. Draw an equilateral triangle PBC on side BC considering BC is a base and an equilateral triangle QAC on digonal AC considering AC is a base. Find the value of\u00a0$\\ \\frac{Area \\ of \\ \\triangle PBC}{Area \\ of \\ \\triangle QAC}$<\/p>\n

a)\u00a0$\\frac{1}{2}$<\/p>\n

b)\u00a01<\/p>\n

c)\u00a0$\\frac{1}{3}$<\/p>\n

d)\u00a0$\\frac{1}{4}$<\/p>\n

SSC CGL Previous Papers Download PDF<\/a><\/p>\n

SSC CGL Free Mock Test<\/a><\/p>\n

Answers & Solutions:<\/strong><\/span><\/p>\n

1)\u00a0Answer\u00a0(C)<\/strong><\/p>\n

<\/p>\n

BC = 500 m<\/p>\n

Let CD be $x$ => BD = $500 – x$<\/p>\n

From $\\triangle$ABC<\/p>\n

=> $tan 30 = \\frac{AB}{BC}$<\/p>\n

=> $\\frac{1}{\\sqrt{3}} = \\frac{AB}{500}$<\/p>\n

=> $AB = \\frac{500}{\\sqrt{3}}$ m<\/p>\n

Now, from $\\triangle$ABD<\/p>\n

=> $tan 60 = \\frac{AB}{BD}$<\/p>\n

=> $\\sqrt{3} = \\frac{\\frac{500}{\\sqrt{3}}}{500 – x}$<\/p>\n

=> $3 (500 – x) = 500$<\/p>\n

=> $3x = 1000$<\/p>\n

$\\therefore x = \\frac{1000}{3}$ metre = $\\frac{1}{3}$ km<\/p>\n

Also, speed of car = $\\frac{distance}{time}$<\/p>\n

= $\\frac{\\frac{1}{3}}{\\frac{10}{60 * 60}}$ km\/hr<\/p>\n

= 120 km\/hr<\/p>\n

2)\u00a0Answer\u00a0(D)<\/strong><\/p>\n

Expression : $\\frac{sin39^{\\circ}}{cos51^{\\circ}}$ + $2 tan11^{\\circ} tan31^{\\circ} tan45^{\\circ} tan59^{\\circ} tan79^{\\circ}$ – $3(sin^{2}21^{\\circ} + sin^{2}69^{\\circ})$<\/p>\n

= $\\frac{sin 39^{\\circ}}{cos (90^{\\circ} – 39^{\\circ})}$ + $2 \\times tan 11^{\\circ} \\times tan 31^{\\circ}$ $\\times 1 \\times tan (90^{\\circ} – 31^{\\circ}) \\times tan (90^{\\circ} – 11^{\\circ})$ – $3 [sin^2 21^{\\circ} + cos^2 (90^{\\circ} – 21^{\\circ})]$<\/p>\n

= $\\frac{sin 39^{\\circ}}{sin 39^{\\circ}}$ + $2 \\times tan 11^{\\circ} \\times cot 11^{\\circ} \\times tan 31^{\\circ} \\times cot 31^{\\circ}$ – $3(sin^2 21^{\\circ} + cos^2 21^{\\circ})$<\/p>\n

= $1 + 2 – 3 = 0$<\/p>\n

3)\u00a0Answer\u00a0(B)<\/strong><\/p>\n

Using section formula, the coordinates of point that divides line joining A = $(x_1 , y_1)$ and B = $(x_2 , y_2)$ in the ratio a\u00a0: b<\/p>\n

= $(\\frac{a x_2 + b x_1}{a + b} , \\frac{a y_2 + b y_1}{a + b})$<\/p>\n

Coordinates of A(0,4) and B(-5,9). Let coordinates of\u00a0P = (x,y) which divides AB in ratio = 2\u00a0: 3<\/p>\n

=> $x = \\frac{(2 \\times -5) + (3 \\times 0)}{2 + 3}$<\/p>\n

=> $5x = -10$<\/p>\n

=> $x = \\frac{-10}{5} = -2$<\/p>\n

Similarly, $y = \\frac{(2 \\times 9) + (3 \\times 4)}{2 + 3}$<\/p>\n

=> $5y = 18 + 12 = 30$<\/p>\n

=> $y = \\frac{30}{5} = 6$<\/p>\n

=> Point P = (-2,6)<\/p>\n

Slope of AB = $\\frac{9 – 4}{-5 – 0} = \\frac{5}{-5} = -1$<\/p>\n

Let slope of line perpendicular to AB = $m$<\/p>\n

Also, product of slopes of two perpendicular lines is -1<\/p>\n

=> $m \\times -1 = -1$<\/p>\n

=> $m = 1$<\/p>\n

Equation of lines having slope $m$ and passing through point P(-2,6) is<\/p>\n

=> $(y – 6) = 1(x + 2)$<\/p>\n

=> $y – 6 = x + 2$<\/p>\n

=> $x – y = -8$<\/p>\n

=> Ans – (B)<\/p>\n

4)\u00a0Answer\u00a0(A)<\/strong><\/p>\n

Total number of students = 40<\/p>\n

Among the student who have passed, Raman’s rank from top = 13th<\/p>\n

Raman’s rank from bottom = 17th<\/p>\n

=> Total students who passed = $(13+17)-1=30-1=29$<\/p>\n

$\\therefore$ Number of students who have failed = $40-29=11$<\/p>\n

=> Ans – (A)<\/p>\n

5)\u00a0Answer\u00a0(A)<\/strong><\/p>\n

Let the number is 100<\/p>\n

First, Increased by 30%\u00a0 =\u00a0 $\\frac{130}{100} \\times 100 = 130$<\/p>\n

Then,\u00a0decreased by 30% =\u00a0$\\frac{70}{100} \\times 130 = 91$<\/p>\n

Then further decreased by 30% =\u00a0$\\frac{70}{100} \\times 91 = 63.7$<\/p>\n

So change = $63.7-100=-36.3$\u00a0 \u00a0or\u00a0 \u00a0 $36.3$ % decrease.<\/p>\n

$\\therefore$ change % = $36$% decrease\u00a0 (nearest integer)<\/p>\n

6)\u00a0Answer\u00a0(D)<\/strong><\/p>\n

Surface area of cuboid=2(8\u00d712+12\u00d710+8\u00d710)=592 .<\/p>\n

Let say it cutted into n number of small cubes.<\/p>\n

Then according to question:<\/p>\n

$n\u00d72^3=8\u00d712\u00d710$<\/p>\n

or, n=120.<\/p>\n

So, Total surface area of all cubes<\/p>\n

$=120\u00d76\u00d72^2$=2880.<\/p>\n

So, percentage increase in Total surface area=((2880-592)\/592)\u00d7100=386.4%.<\/p>\n

D is correct choice.<\/p>\n

7)\u00a0Answer\u00a0(C)<\/strong><\/p>\n

Marked price = Rs. 18,000<\/p>\n

Selling price after first discount of 10% = $18,000-(\\frac{10}{100}\\times18,000)$<\/p>\n

=\u00a0$18,000-1800=Rs.$ $16,200$<\/p>\n

Similarly, selling price after second discount of 5% = $16,200-(\\frac{5}{100}\\times16,200)$<\/p>\n

=\u00a0$16,200-810=Rs.$ $15,390$<\/p>\n

And selling price after third discount of 2% = $15,390-(\\frac{2}{100}\\times15,390)$<\/p>\n

=\u00a0$15,390-307.8=Rs.$ $15,082.20$<\/p>\n

=> Ans – (C)<\/p>\n

8)\u00a0Answer\u00a0(C)<\/strong><\/p>\n

Let\u00a0the number of students who passed the examination = $x$<\/p>\n

=> Number of students who failed = $(150-x)$<\/p>\n

Average marks of 150 students = 40<\/p>\n

=> Total marks = $40\\times150=6000$<\/p>\n

Similarly, total marks scored by passed students = $60x$<\/p>\n

According to ques,<\/p>\n

=> $60x+20(150-x)=6000$<\/p>\n

=> $60x+3000-20x=6000$<\/p>\n

=> $40x=6000-3000=3000$<\/p>\n

=> $x=\\frac{3000}{40}=75$<\/p>\n

$\\therefore$ The number of students who passed the examination =\u00a075<\/strong><\/p>\n

=> Ans – (C)<\/p>\n

9)\u00a0Answer\u00a0(A)<\/strong><\/p>\n

Ratio of milk to water in solution 1 is 3:2
\nSolution 1 contains 40 liters.
\nMilk content in solution 1 =(3\/5)*40
\n=24 liters
\nWater content=40-24=16 liters
\nRatio of milk to water in solution 2 is 2:1
\nSolution 2 contains 30 liters.
\nMilk content in solution 2 =(2\/3)*30
\n=20 liters
\nWater content=30-20=10 liters
\nTotal milk content=24+20=44 liters
\nTotal water content =16+10=26 liters
\nRatio=22:13<\/p>\n

10)\u00a0Answer\u00a0(A)<\/strong><\/p>\n

Investment of Teena = Rs.24000
\nDuration of Teena = 8 months
\nTotal investment of Teena = $Rs.24000 \\times 8$
\nInvestment of Reena = Rs.28000
\nDuration of Reena = 10 months
\nTotal investment of Reena = $Rs.28000\\times10$
\nInvestment of Sheena = Rs.20000
\nDuration of Sheena = 12 months
\nTotal investment of Sheena = $Rs.20000\\times12$
\nRatio of profits = Ratio of investments =\u00a0$Rs.24000 \\times 8 :\u00a0Rs.28000\\times10 :\u00a0Rs.20000\\times12 = 24 :\u00a035 : 30$
\nGiven, Total profit = Rs.25810
\nThen, Profit of Teena = $\\dfrac{24}{89} \\times 25810 = Rs.6960$<\/p>\n

11)\u00a0Answer\u00a0(B)<\/strong><\/p>\n

Let s and u be the no. of candidates selected and unselected respectively.<\/p>\n

Total candidates applied = s+u<\/p>\n

Condition 1:<\/strong><\/p>\n

Ratio of selected to unselected was 19: 17.<\/em><\/p>\n

s\/u = (19\/17)<\/p>\n

17s= 19u \u2014\u2014(1)<\/p>\n

Condition 2:<\/strong><\/p>\n

If 1200 less had applied and 800 less selected, the ratio of selected to unselected would have been 1 : 1<\/em><\/p>\n

Total candidates applied = s+u-1200<\/p>\n

Selected candidates = s-800<\/p>\n

Then, unselected candidates = s+u-1200 – (s-800) = s+u-1200-s+800 = u-400<\/p>\n

(s-800)\/(u-400) = (1\/1)<\/p>\n

s-800 = u-400 \u2014-(2)<\/p>\n

From (1),<\/p>\n

u = (17\/19)s<\/p>\n

Substitute in (2),<\/p>\n

s-800 = (17\/19)s -400<\/p>\n

s= 3800<\/p>\n

u = 3400<\/p>\n

s+u = 3800 + 3400 = 7200<\/p>\n

The number of applicants is 7200<\/strong><\/p>\n

B is correct choice.
\n<\/strong><\/p>\n

12)\u00a0Answer\u00a0(D)<\/strong><\/p>\n

As we know that<\/p>\n

Speed = $\\frac{distance}{time}$<\/p>\n

Given<\/p>\n

Let train length = D<\/p>\n

Speed = $\\frac{D}{60}$<\/p>\n

Now<\/p>\n

Given that<\/p>\n

Bridge length = 2D<\/p>\n

Time = $\\frac{bridge length+ train }{speed}$<\/p>\n

Time = $\\frac{2D+ D }{\\frac{D}{60}}$<\/p>\n

Time = 180<\/p>\n

13)\u00a0Answer\u00a0(D)<\/strong><\/p>\n

P \u00a0= 7 hrs<\/p>\n

Q =13 hrs<\/p>\n

Total work = lcm (7,13)= 91 hrs<\/p>\n

P’s effiency = $91\\div7$ =13<\/p>\n

Q’s effiency = $91\\div13$ =7<\/p>\n

(P+Q)’s together takes\u00a0 = $Total work\\div total efficency$<\/p>\n

= $91\\div20$<\/p>\n

convert 91 hrs into minutes = $91\\times 60=5460$<\/p>\n

= $5460\\div20$<\/p>\n

= 273 min<\/p>\n

=\u00a0$273\\div60$<\/p>\n

=4 hours 33 minutes
\n14)\u00a0Answer\u00a0(A)<\/strong><\/p>\n

Let the speed of the train from Q be X km\/hr
\nThen the speed of the train from P is ( X+8 ) km\/hr
\nThey meet each other at M after 6 hours travel
\nPM + MQ = 162 km
\n6 ( X + 8 ) + 6X = 162 km
\n12X + 48 = 162
\n12X = 162 – 48
\n12X = 114
\nX = 114\/12
\n9 (1\/2) km\/hr<\/p>\n

15)\u00a0Answer\u00a0(A)<\/strong><\/p>\n

<\/figure>\n

Let side of the square be $x$ cm<\/p>\n

=> Side of equilateral $\\triangle$ PBC = $x$ cm<\/p>\n

In right $\\triangle$ ABC,<\/p>\n

=> $(AC)^2=(AB)^2+(BC)^2$<\/p>\n

=> $(AC)^2=(x)^2+(x)^2=2x^2$<\/p>\n

=> $AC=\\sqrt2x$<\/p>\n

$\\therefore$\u00a0$\\ \\frac{ar(\\triangle PBC)}{ar(\\triangle QAC)}$<\/p>\n

= $[\\frac{\\sqrt3}{4}\\times (x)^2]\\div[\\frac{\\sqrt3}{4}\\times\u00a0(\\sqrt2x)^2]$<\/p>\n

= $\\frac{x^2}{2x^2}=\\frac{1}{2}$<\/p>\n

=> Ans – (A)<\/p>\n

Get 200 SSC mocks for just Rs. 249. Enroll here<\/a><\/p>\n

SSC Free Previous Papers App<\/a><\/p>\n

We hope this Advance Maths Questions for SSC CGL Exams will be highly useful for your preparation.<\/p>\n","protected":false},"excerpt":{"rendered":"

Advance Maths Questions for SSC CGL PDF Download SSC CGL Advance Maths Questions PDF based on previous papers very useful for SSC CGL Exams. Top-15 Very Important Maths Questions for SSC CGL Exam. Question 1:\u00a0A car is travelling on a straight road leading to a tower. From a point at a distance of 500 m […]<\/p>\n","protected":false},"author":41,"featured_media":41000,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"om_disable_all_campaigns":false,"_mi_skip_tracking":false,"footnotes":""},"categories":[169,125,9,504,378,1441],"tags":[165,462,358,3043],"class_list":{"0":"post-40998","1":"post","2":"type-post","3":"status-publish","4":"format-standard","5":"has-post-thumbnail","7":"category-downloads","8":"category-featured","9":"category-ssc","10":"category-ssc-cgl","11":"category-ssc-chsl","12":"category-stenographer","13":"tag-ssc","14":"tag-ssc-cgl","15":"tag-ssc-chsl","16":"tag-ssc-maths-questions"},"better_featured_image":{"id":41000,"alt_text":"Advance Maths Questions for SSC CGL PDF","caption":"Advance Maths Questions for SSC CGL PDF","description":"Advance Maths Questions for SSC CGL PDF","media_type":"image","media_details":{"width":1200,"height":630,"file":"2020\/03\/fig-05-03-2020_05-39-33.jpg","sizes":{"medium":{"file":"fig-05-03-2020_05-39-33-300x158.jpg","width":300,"height":158,"mime-type":"image\/jpeg","source_url":"https:\/\/cracku.in\/blog\/wp-content\/uploads\/2020\/03\/fig-05-03-2020_05-39-33-300x158.jpg"},"large":{"file":"fig-05-03-2020_05-39-33-1024x538.jpg","width":1024,"height":538,"mime-type":"image\/jpeg","source_url":"https:\/\/cracku.in\/blog\/wp-content\/uploads\/2020\/03\/fig-05-03-2020_05-39-33-1024x538.jpg"},"thumbnail":{"file":"fig-05-03-2020_05-39-33-150x150.jpg","width":150,"height":150,"mime-type":"image\/jpeg","source_url":"https:\/\/cracku.in\/blog\/wp-content\/uploads\/2020\/03\/fig-05-03-2020_05-39-33-150x150.jpg"},"medium_large":{"file":"fig-05-03-2020_05-39-33-768x403.jpg","width":768,"height":403,"mime-type":"image\/jpeg","source_url":"https:\/\/cracku.in\/blog\/wp-content\/uploads\/2020\/03\/fig-05-03-2020_05-39-33-768x403.jpg"},"tiny-lazy":{"file":"fig-05-03-2020_05-39-33-30x16.jpg","width":30,"height":16,"mime-type":"image\/jpeg","source_url":"https:\/\/cracku.in\/blog\/wp-content\/uploads\/2020\/03\/fig-05-03-2020_05-39-33-30x16.jpg"},"td_218x150":{"file":"fig-05-03-2020_05-39-33-218x150.jpg","width":218,"height":150,"mime-type":"image\/jpeg","source_url":"https:\/\/cracku.in\/blog\/wp-content\/uploads\/2020\/03\/fig-05-03-2020_05-39-33-218x150.jpg"},"td_324x400":{"file":"fig-05-03-2020_05-39-33-324x400.jpg","width":324,"height":400,"mime-type":"image\/jpeg","source_url":"https:\/\/cracku.in\/blog\/wp-content\/uploads\/2020\/03\/fig-05-03-2020_05-39-33-324x400.jpg"},"td_696x0":{"file":"fig-05-03-2020_05-39-33-696x365.jpg","width":696,"height":365,"mime-type":"image\/jpeg","source_url":"https:\/\/cracku.in\/blog\/wp-content\/uploads\/2020\/03\/fig-05-03-2020_05-39-33-696x365.jpg"},"td_1068x0":{"file":"fig-05-03-2020_05-39-33-1068x561.jpg","width":1068,"height":561,"mime-type":"image\/jpeg","source_url":"https:\/\/cracku.in\/blog\/wp-content\/uploads\/2020\/03\/fig-05-03-2020_05-39-33-1068x561.jpg"},"td_0x420":{"file":"fig-05-03-2020_05-39-33-800x420.jpg","width":800,"height":420,"mime-type":"image\/jpeg","source_url":"https:\/\/cracku.in\/blog\/wp-content\/uploads\/2020\/03\/fig-05-03-2020_05-39-33-800x420.jpg"},"td_80x60":{"file":"fig-05-03-2020_05-39-33-80x60.jpg","width":80,"height":60,"mime-type":"image\/jpeg","source_url":"https:\/\/cracku.in\/blog\/wp-content\/uploads\/2020\/03\/fig-05-03-2020_05-39-33-80x60.jpg"},"td_100x70":{"file":"fig-05-03-2020_05-39-33-100x70.jpg","width":100,"height":70,"mime-type":"image\/jpeg","source_url":"https:\/\/cracku.in\/blog\/wp-content\/uploads\/2020\/03\/fig-05-03-2020_05-39-33-100x70.jpg"},"td_265x198":{"file":"fig-05-03-2020_05-39-33-265x198.jpg","width":265,"height":198,"mime-type":"image\/jpeg","source_url":"https:\/\/cracku.in\/blog\/wp-content\/uploads\/2020\/03\/fig-05-03-2020_05-39-33-265x198.jpg"},"td_324x160":{"file":"fig-05-03-2020_05-39-33-324x160.jpg","width":324,"height":160,"mime-type":"image\/jpeg","source_url":"https:\/\/cracku.in\/blog\/wp-content\/uploads\/2020\/03\/fig-05-03-2020_05-39-33-324x160.jpg"},"td_324x235":{"file":"fig-05-03-2020_05-39-33-324x235.jpg","width":324,"height":235,"mime-type":"image\/jpeg","source_url":"https:\/\/cracku.in\/blog\/wp-content\/uploads\/2020\/03\/fig-05-03-2020_05-39-33-324x235.jpg"},"td_356x220":{"file":"fig-05-03-2020_05-39-33-356x220.jpg","width":356,"height":220,"mime-type":"image\/jpeg","source_url":"https:\/\/cracku.in\/blog\/wp-content\/uploads\/2020\/03\/fig-05-03-2020_05-39-33-356x220.jpg"},"td_356x364":{"file":"fig-05-03-2020_05-39-33-356x364.jpg","width":356,"height":364,"mime-type":"image\/jpeg","source_url":"https:\/\/cracku.in\/blog\/wp-content\/uploads\/2020\/03\/fig-05-03-2020_05-39-33-356x364.jpg"},"td_533x261":{"file":"fig-05-03-2020_05-39-33-533x261.jpg","width":533,"height":261,"mime-type":"image\/jpeg","source_url":"https:\/\/cracku.in\/blog\/wp-content\/uploads\/2020\/03\/fig-05-03-2020_05-39-33-533x261.jpg"},"td_534x462":{"file":"fig-05-03-2020_05-39-33-534x462.jpg","width":534,"height":462,"mime-type":"image\/jpeg","source_url":"https:\/\/cracku.in\/blog\/wp-content\/uploads\/2020\/03\/fig-05-03-2020_05-39-33-534x462.jpg"},"td_696x385":{"file":"fig-05-03-2020_05-39-33-696x385.jpg","width":696,"height":385,"mime-type":"image\/jpeg","source_url":"https:\/\/cracku.in\/blog\/wp-content\/uploads\/2020\/03\/fig-05-03-2020_05-39-33-696x385.jpg"},"td_741x486":{"file":"fig-05-03-2020_05-39-33-741x486.jpg","width":741,"height":486,"mime-type":"image\/jpeg","source_url":"https:\/\/cracku.in\/blog\/wp-content\/uploads\/2020\/03\/fig-05-03-2020_05-39-33-741x486.jpg"},"td_1068x580":{"file":"fig-05-03-2020_05-39-33-1068x580.jpg","width":1068,"height":580,"mime-type":"image\/jpeg","source_url":"https:\/\/cracku.in\/blog\/wp-content\/uploads\/2020\/03\/fig-05-03-2020_05-39-33-1068x580.jpg"}},"image_meta":{"aperture":"0","credit":"","camera":"","caption":"","created_timestamp":"0","copyright":"","focal_length":"0","iso":"0","shutter_speed":"0","title":"","orientation":"0"}},"post":null,"source_url":"https:\/\/cracku.in\/blog\/wp-content\/uploads\/2020\/03\/fig-05-03-2020_05-39-33.jpg"},"yoast_head":"\n\n\n\n\n\n\n\n\n\n\n\n\n\n\t\n\t\n\n\n\n